4
$\begingroup$

We know that an $m \times m$ random matrix $\boldsymbol{A} = \boldsymbol{H} \boldsymbol{H}^H$ is a (central) real/complex Wishart matrix with $n$ degrees of freedom and covariance matrix $\boldsymbol{\Sigma}$ $(\boldsymbol{A} \sim \mathcal{W}_m (n, \boldsymbol{\Sigma}))$, if the columns of the $m \times n$ matrix $\boldsymbol{H}$ are zero mean independent real/complex Gaussian vectors with covariance matrix $\boldsymbol{\Sigma}$.

If the columns of matrix $\boldsymbol{H}$ are zero mean independent real/complex Gaussian vectors with different covariance matrix, i.e, the $i$-th column of $\boldsymbol{H}$ has covariance matrix $\boldsymbol{\Sigma}_i$ ($i=1, \cdots, n$), is matrix $\boldsymbol{A}$ a Wishart matrix, and what is its distribution? Thanks!

| cite | improve this question | |
$\endgroup$
3
$\begingroup$

It is not a Wishart random matrix. When $m=1$ then ${\cal W}_1(n,\sigma^2) = \sigma^2 \chi^2_n= \Gamma\left(\frac{n}{2},\frac{1}{2\sigma^2}\right)$ is the distribution of $\sum_{i=1}^n {X_i}^2$ where $X_i\sim_{\text{iid}} {\cal N}(0,\sigma^2)$.

Taking $X_i\sim{\cal N}(0,\sigma_ i^2)$ with different $\sigma_i$'s yields a sum of independent Gamma distributions with different scale parameters, and this is not a Gamma distribution in general.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for your help! Does your answer mean that when $\boldsymbol{A}$ is not a Wishart matrix when $m = 1$, thus it is not a Wishart matrix when $m > 1$? $\endgroup$ – lala Jul 10 '14 at 8:12
  • $\begingroup$ Because $\boldsymbol{A}$ is a really matrix in my problem ($m > 1$), so does it have any way to prove in general case? Thanks $\endgroup$ – lala Jul 10 '14 at 8:18
  • $\begingroup$ @lala My answer shows this is not true for $m=1$. For $m>1$, one can show that the same fact occurs for the diagonal entries of the matrix: the diagonal entries of a random Wishart matrix are Gamma random variables, but they are not Gamma anymore if you use different $\Sigma$'s. $\endgroup$ – Stéphane Laurent Jul 10 '14 at 9:07
  • $\begingroup$ @ Stéphen Laurent: Could I ask you one more question? How can I prove the diagonal entries of matrix $\boldsymbol{A}$ are not Gamma random variables if using the different $\boldsymbol{\Sigma}$? Must I use the pdf? (It is difficult way, isn't it?) $\endgroup$ – lala Jul 10 '14 at 11:02
  • $\begingroup$ @lala Consider the $(1,1)$-entry of the random matrix. If you take different values of the $(1,1)$-entry of $\Sigma$, then this is exactly the situation when $m=1$. $\endgroup$ – Stéphane Laurent Jul 10 '14 at 11:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.